Show that $(k!)!$ is divisible by $(k!)^{(k-1)!}$

Question

Question : Show that $(k!)!$ is divisible by $(k!)^{(k-1)!}$.

Answer : Suppose we've $k!$ objects, where k objects of 1st kind, k objects of 2nd kind,...k objects of $(k-1)!$ kind (why?). And from the well known formula we get :

$$ \frac{(k!)!}{k!k!...k!((k-1) times)**(why?)**} = \frac{(k!)!}{(k!)^{(k-1)!}}$$

My question is from where we do we get the idea of having $(k-1)$ in the calculation. I'm confused. Please help me.

Answer 1

Let $\lfloor r\rfloor$ denote the floor of $r$
(i.e. the largest integer $\leq r)$.

Lemma 1
For $a \in \Bbb{R}_{\geq 0}, ~b\in \Bbb{Z^+}, ~ b\lfloor a\rfloor \leq \lfloor ab\rfloor.$

Proof:
Let $a = P + r,~~$ where $P \in \Bbb{Z}_{\geq 0}, ~~0 \leq r < 1.$
Thus, $P = \lfloor a\rfloor$.
Further, $(ba) - ( bP) = (br).$
Let $c = \lfloor (br) \rfloor \implies 0 \leq c.$

Then, $\lfloor (ba) \rfloor = (bP) + c \implies \lfloor (ba) \rfloor \geq (bP) = b\lfloor a\rfloor.$

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Given two positive integers, $a,b$, you can show that $\frac{a}{b}$ is an integer by showing that for any prime $p$, if $p^r$ is in the prime factorization of $b$, then $p^s$ is in the prime factorization of $a$, where $s \geq r.$

Use Legendre formula:

$$v_p(n!) = \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k}\right\rfloor.$$

In order to show that $(k!)^{(k-1)!}$ divides $[(k!)!]$, it is sufficient to show that for any prime $p$, if $d$ is the largest non-negative integer such that $p^d$ divides $(k!)$, then $p^{d(k-1)!}$ divides $[(k!)!].$

So, by assumption:

$$ \left\lfloor \frac{k}{p^1}\right\rfloor ~+~ \left\lfloor \frac{k}{p^2}\right\rfloor ~+~ \left\lfloor \frac{k}{p^3}\right\rfloor ~+~ \cdots = d.\tag1$$

It is sufficient to prove that :

$$ \left\lfloor \frac{k!}{p^1}\right\rfloor ~+~ \left\lfloor \frac{k!}{p^2}\right\rfloor ~+~ \left\lfloor \frac{k!}{p^3}\right\rfloor ~+~ \cdots \geq d(k-1)!.\tag1$$

However, this result is immediate, by invoking Lemma 1, term by term.

That is:

$$\left\lfloor \frac{k}{p^1}\right\rfloor \times (k-1)! \leq \left\lfloor \frac{k!}{p^1}\right\rfloor.$$

$$\left\lfloor \frac{k}{p^2}\right\rfloor \times (k-1)! \leq \left\lfloor \frac{k!}{p^2}\right\rfloor.$$

$$\left\lfloor \frac{k}{p^3}\right\rfloor \times (k-1)! \leq \left\lfloor \frac{k!}{p^3}\right\rfloor.$$

$$\cdots$$